Q15 - Triangle median problem

Question:
In a triangle ABC, the median from B to CA is perpendicular to the median from C to AB. If the median from A to BC is 30, determine $$ \frac{BC^2 + CA^2 + AB^2}{100}. $$



 Answer: 24 $$ AR = 30 $$ $$ AG : GR = 2:1 \\ $$ $$ GR = 10 \\ $$ $$ \text{in } \triangle BGC, \angle G = 90^\circ \\ $$ $$ BR = RC = GR = 10 \\ $$ $$ AB^2 + AC^2 = 2 (AR^2 + BR^2) \\ $$ $$ AB^2 + AC^2 = 2 (900 + 100) \\ $$ $$ AB^2 + AC^2 = 2000 \\ $$ $$ \frac{AB^2 + AC^2 + BC^2}{100} = \frac{2000 + 400}{100} = 24 $$

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